The pH of a solution of 4.79 mol/L acetic acid

Question

The pH of a solution of 4.79 mol/L acetic acid in water is a.bc

Answer 1

Acetic acid = CH3COOH

Ka = 1.8x10^-5 (look it up)

CH3COOH ==> CH3COO^- + H⁺

Ka = [H+][CH3COO-] / [CH3COOH]

1.8x10^-5 = (x)(x) / 4.79 - x (assume x in denominator is small and ignore it)

x² = 8.62x10^-5 (note that this is small compared to 479 so assumption made above is valid)

x = 9.29x10^-3 mol/L = [H+]

pH = -log [H⁺]

pH = -log 9.29x10^-3

pH = 2.03

a = 2

b = 0

c = 3